The equation of a circle $C$ is $x^2+y^2+12x-4y+31 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+12x) + (y^2-4y) = -31$ $(x^2+12x+36) + (y^2-4y+4) = -31 + 36 + 4$ $(x+6)^{2} + (y-2)^{2} = 9 = 3^2$ Thus, $(h, k) = (-6, 2)$ and $r = 3$.